Functions of a Matrix

The motivation is to study matrix-valued functions, which stems from the differential equations describing linear systems.

$$\large \dot{x} = A x(t)$$

$$\large x(t) = e^{At} x(0)$$

The power series representation of any function is

$$\large f(s) = \sum_{i=0}^{\infty} \alpha_i s^i$$

where $s \in \mathbb{C}$.

The power series representation of a matrix-valued function is

$$\large f(A) = \sum_{i=0}^{\infty} \alpha_iA^i$$

where $A \in \mathbb{C}^{n \times n}$. This solution is another matrix as the same size as $A$.

The exponential function is defined for matrices as

$$\large e^t = \sum_{i=0}^{\infty} \frac{t^i}{i!} \implies e^A = \sum_{i=0}^{\infty} \frac{A^i}{i!}$$

By using Cayley-Hamilton Theorem, we can write $A^i$ as a linear combination of $I,A,A^2,\cdots,A^{n-1}$.

$$\large e^A = c_0 I + c_1 A + c_2 A^2 + \cdots + c_{n-1} A^{n-1}$$

where $c_i$ are scalars.

Remark: One can use the minimal polynomial of a matrix to express the $l$th power of a matrix in terms of $I,A,A^2,\cdots,A^{l-1}$. $l$ is the order of the minimal polynomial.

$$\large e^A = c_0 I + c_1 A + c_2 A^2 + \cdots + c_{l-1} A^{l-1}$$

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First Method

Let

$$f(s) = \sum_{i=0}^{\infty} \alpha_i s^i$$

$$f(A) = \sum_{i=0}^{\infty} \alpha_i A^i$$

Define $p(s)$ and $P(A)$ as follows

$$\large p(s) = c_0 + c_1 s + c_2 s^2 + \cdots + c_{l-1} s^{l-1}$$

$$\large P(A) = c_0 I + c_1 A + c_2 A^2 + \cdots + c_{l-1} A^{l-1}$$

Then we have the equality

$$\large f(A) = P(A)$$

where $P(A)$ is a polynomial of $A$.

Case I: $A$ is diagonalizable

Suppose

$\large m(s) = (s - \lambda_1 ) (s - \lambda_2 ) \cdots (s - \lambda_{\sigma} )$

$l = \sigma$ , $m_1 = m_2 = \cdots = m_{\sigma} = 1$

$\large f(A) = P(A) = c_0 I + c_1 A + c_2 A^2 + \cdots + c_{l-1} A^{l-1}$

$Ae_i = \lambda_i e_i$

Let $e_i$ be the eigenvector of $A$ corresponding to $\lambda_i$, multiply both sides by $e_i$.

$\sum_{n=0}^{l-1} \alpha_i A^n e_i = \sum_{n=0}^{l-1} c_n \lambda_i^n e_i$

$\sum_{n=0}^{l-1} \alpha_i \lambda_i^n e_i = \sum_{n=0}^{l-1} c_n \lambda_i^n e_i$

$\alpha_i \lambda_i^n = c_n \lambda_i^n$

$\alpha_i = c_n$

$\large f(A) = P(A) = \sum_{i=0}^{l-1} \alpha_i A^i = \sum_{i=0}^{l-1} c_i A^i$

$\large f(\lambda_i) = P(\lambda_i)$

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Example: $A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$ find $e^A$ and $log(A)$

Solution:

$d(s) = (s - 3)(s - 1)$

$\lambda_1 = 3$, $\lambda_2 = 1$

$m(s) = (s - 3)(s - 1)$, $l = 2$

$p(s) = c_0 + c_1 s$, and $P(A) = c_0 I + c_1 A$

$Ae_1 = 3e_1$, $Ae_2 = e_2$

$f(3) = p(3) \\ f(1) = p(1)$

$f(3) = e^3 = c_0 + 3c_1 \\ f(1) = \ e \ = c_0 + c_1$

$c_1 = \frac{e^3 - e}{2} \\ c_0 = \frac{3e - e^3}{2}$

$e^A = \frac{3e - e^3}{2} I + \frac{e^3 - e}{2} A$

$e^A = \frac{3e - e^3}{2} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \frac{e^3 - e}{2} \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$

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Case II: $A$ is not diagonalizable

Consider the following example. Let $A \in \mathbb{R}^3$ and $m(s) = (s-\lambda_1)^2(s-\lambda_2)$.

$$\large J = \begin{bmatrix} \lambda_1 & 1 & 0 \\ 0 & \lambda_1 & 0 \\ 0 & 0 & \lambda_2 \end{bmatrix}$$

$$\sum m_i = 2 + 1 = 3$$

$$f(A) = P(A) = c_0 I + c_1 A + c_2 A^2$$

$$f(\lambda_1) = P(\lambda_1) \ \ \ f(\lambda_2) = P(\lambda_2)$$

These two equations are not enough to find $c_0,c_1,c_2$.

Consider the matrix $P$ that transforms $A$ into its Jordan canonical form $J$.

We know that $P^{-1} = P^*$ and $P^*AP = J$, and $P$ is in the form of $[e_1,f_1,e_2]$. Where $e_1$ and $e_2$ are the eigenvectors of $A$ corresponding to $\lambda_1$ and $\lambda_2$ respectively. $f_1$ is the generalized eigenvector of $A$ corresponding to $\lambda_1$.

$$\large P = \begin{bmatrix} \vdots & \vdots & \vdots \\ e_1 & f_1 & e_2 \\ \vdots & \vdots & \vdots \end{bmatrix}$$

$$\large \begin{bmatrix} \vdots & \vdots & \vdots \\ e_1 & f_1 & e_2 \\ \vdots & \vdots & \vdots \end{bmatrix} \begin{bmatrix} \lambda_1 & 1 & 0 \\ 0 & \lambda_1 & 0 \\ 0 & 0 & \lambda_2 \end{bmatrix} = A \begin{bmatrix} \vdots & \vdots & \vdots \\ e_1 & f_1 & e_2 \\ \vdots & \vdots & \vdots \end{bmatrix}$$

$$\begin{align*} \large Af_1 \ & = \lambda_1 f_1 + e_1 \\ \large A^2f_1 & = \lambda_1 Af_1 + Ae_1 \\ \large & = \lambda_1^2 f_1 + \lambda_1 e_1 + \lambda_1 e_1 = \lambda_1^2 f_1 + 2 \lambda_1 e_1 \\ \large A^3f_1 & = \lambda_1^2 Af_1 + 2\lambda_1 Ae_1 = \lambda_1^3 f_1 + 3 \large \lambda_1^2 e_1 \\ & \ \ \ \vdots \\ \large A^kf_1 & = \lambda_1^k f_1 + k \lambda_1^{k-1} e_1 \\ \end{align*}$$

Return to the equation $f(A) = P(A)$.

$$\large \sum_{i=0}^{l-1} \alpha_i A^i = \sum_{i=0}^{l-1} c_i A^i$$

Multiply both sides by $f_1$ from the right.

$$\large \sum_{i=0}^{l-1} \alpha_i A^i f_1 = \sum_{i=0}^{l-1} c_i A^i f_1$$

$$\large \sum_{i=0}^{l-1} \alpha_i \lambda_1^i f_1 + \sum_{i=0}^{l-1} \alpha_i i \lambda_1^{i-1} e_1 = \sum_{i=0}^{l-1} c_i \lambda_1^i f_1 + \sum_{i=0}^{l-1} c_i i \lambda_1^{i-1} e_1$$

$$\large f(\lambda_1) f_1 + f'(\lambda_1) e_1 = P(\lambda_1) f_1 + P'(\lambda_1) e_1$$

$$\large f(\lambda_1) = P(\lambda_1)$$

$$\large f'(\lambda_1) = P'(\lambda_1)$$

Which is the additional equation we need to find $c_0,c_1,c_2$.

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General Case

Let $m(s) = (s-\lambda_1)^{m_1}(s-\lambda_2)^{m_2} \cdots (s-\lambda_{\sigma})^{m_{\sigma}}$. We have the following set of equations.

$$\large f^{(t)}(\lambda_i) = P^{(t)}(\lambda_i)$$

where $t = 0,1,2,\cdots,m_i-1$ and $i = 1,2,\cdots,\sigma$. We have $\sum_{i=1}^{\sigma} m_i = l$ equations.

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Example: $A = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}$ Find $\sin{(\pi A)}$.

Solution:

$m(s) = s^3(s-1)^2$ and the order of the total minimal polynomial is $l = 5$.

$p(s) = c_0 + c_1 s + c_2 s^2 + c_3 s^3 + c_4 s^4$

$p(A) = c_0 I + c_1 A + c_2 A^2 + c_3 A^3 + c_4 A^4$

We need five equations to find $c_0,c_1,c_2,c_3,c_4$.

Starting with

$$\begin{align*} f(s) &= \sin{(\pi s)} \\ f'(s) &= \pi \cos{(\pi s)} \\ f''(s) &= -\pi^2 \sin{(\pi s)} \end{align*}$$

Then

$$\begin{align*} p(s) &= c_0 + c_1 s + c_2 s^2 + c_3 s^3 + c_4 s^4 \\ p'(s) &= c_1 + 2c_2 s + 3c_3 s^2 + 4c_4 s^3 \\ p''(s) &= 2c_2 + 6c_3 s + 12c_4 s^2\end{align*}$$

For the first eigenvalue $\lambda_1 = 0$. We have $m_1 = 3$.

$$\begin{align*} & f(\lambda_1) = p(\lambda_1) \\ & f'(\lambda_1) = p'(\lambda_1) \\ & f''(\lambda_1) = p''(\lambda_1) \end{align*}$$

$$\begin{align*} & f(0) = p(0) \\ & f'(0) = p'(0) \\ & f''(0) = p''(0) \end{align*}$$

$$\begin{align*} & \sin{(0)} = 0 = c_0 \\ & \pi \cos{(0)} = \pi = c_1 \\ & -\pi^2 \sin{(0)} = 0 = 2c_2 \end{align*}$$

For the second eigenvalue $\lambda_2 = 1$. We have $m_2 = 2$.

$$\begin{align*} & f(\lambda_2) = p(\lambda_2) \\ & f'(\lambda_2) = p'(\lambda_2) \end{align*}$$

$$\begin{align*} & \sin{(\pi)} = 0 = c_0 + c_1 + c_2 + c_3 + c_4 \\ & \pi \cos{(\pi)} = -\pi = c_1 + 2c_2 + 3c_3 + 4c_4 \end{align*}$$

$$\large \begin{bmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \\ c_4 \end{bmatrix} = \begin{bmatrix} 0 \\ \pi \\ 0 \\ -\pi \\ -\pi \end{bmatrix}$$

$$\large \begin{bmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \\ c_4 \end{bmatrix} = \begin{bmatrix} 0 \\ \pi \\ 0 \\ -2\pi \\ \pi \end{bmatrix}$$

$$\large p(s) = \pi s - 2\pi s^3 + \pi s^4$$

$$\large \sin{(\pi A)} = \pi A - 2\pi A^3 + \pi A^4$$

Remark: $f(A)$ does not exist when $f^{(t)}(\lambda_i)$ does not exist for some $t$ and $i$.

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#EE501 - Linear Systems Theory at METU